Answer :
To determine the Gibbs free energy ([tex]\(\Delta G\)[/tex]) of the reaction given the standard enthalpy change ([tex]\(\Delta H^0\)[/tex]) and the standard entropy change ([tex]\(\Delta S^0\)[/tex]) at a specific temperature (T), you can use the Gibbs free energy equation:
[tex]\[ \Delta G = \Delta H - T\Delta S \][/tex]
Here are the provided values:
- Temperature (T) = 298 K
- Enthalpy change ([tex]\(\Delta H^0\)[/tex]) = -1652 kJ/mol
- Entropy change ([tex]\(\Delta S^0\)[/tex]) = 0.097 kJ/(K·mol)
Follow these steps for calculation:
1. Write down the equation for Gibbs free energy:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
2. Substitute the given values into the equation:
[tex]\[ \Delta G = -1652 \, \text{kJ/mol} - 298 \, \text{K} \times 0.097 \, \text{kJ/(K·mol)} \][/tex]
3. Perform the multiplication [tex]\( T \Delta S \)[/tex]:
[tex]\[ 298 \, \text{K} \times 0.097 \, \text{kJ/(K·mol)} = 28.906 \, \text{kJ/mol} \][/tex]
4. Subtract this result from [tex]\(\Delta H^0\)[/tex]:
[tex]\[ \Delta G = -1652 \, \text{kJ/mol} - 28.906 \, \text{kJ/mol} = -1680.906 \, \text{kJ/mol} \][/tex]
Therefore, the Gibbs free energy ([tex]\(\Delta G\)[/tex]) of the reaction at 298 K is:
[tex]\[ -1680.906 \, \text{kJ/mol} \][/tex]
None of the provided options (A, B, C, D) match [tex]\(-1680.906 \, \text{kJ/mol}\)[/tex]. Therefore, there might be a discrepancy between the provided options and the correct computed value.
[tex]\[ \Delta G = \Delta H - T\Delta S \][/tex]
Here are the provided values:
- Temperature (T) = 298 K
- Enthalpy change ([tex]\(\Delta H^0\)[/tex]) = -1652 kJ/mol
- Entropy change ([tex]\(\Delta S^0\)[/tex]) = 0.097 kJ/(K·mol)
Follow these steps for calculation:
1. Write down the equation for Gibbs free energy:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
2. Substitute the given values into the equation:
[tex]\[ \Delta G = -1652 \, \text{kJ/mol} - 298 \, \text{K} \times 0.097 \, \text{kJ/(K·mol)} \][/tex]
3. Perform the multiplication [tex]\( T \Delta S \)[/tex]:
[tex]\[ 298 \, \text{K} \times 0.097 \, \text{kJ/(K·mol)} = 28.906 \, \text{kJ/mol} \][/tex]
4. Subtract this result from [tex]\(\Delta H^0\)[/tex]:
[tex]\[ \Delta G = -1652 \, \text{kJ/mol} - 28.906 \, \text{kJ/mol} = -1680.906 \, \text{kJ/mol} \][/tex]
Therefore, the Gibbs free energy ([tex]\(\Delta G\)[/tex]) of the reaction at 298 K is:
[tex]\[ -1680.906 \, \text{kJ/mol} \][/tex]
None of the provided options (A, B, C, D) match [tex]\(-1680.906 \, \text{kJ/mol}\)[/tex]. Therefore, there might be a discrepancy between the provided options and the correct computed value.
